Question

How do you solve for vertical asymptote? Like, I know you set the denominator to 0, but after that I'm confused.
the function is f(x) = (3x+1)/(2x)

Answer 1

2x=0
x=0

Answer 2

There is a vertical asymptote at x=0

Answer 3

so, x=0 is a line, the y-axis which is the vertical asymptote to your function f(x)

Answer 4

2x = 0 => x = 0

those are just the vertical asymptotes though

Answer 5

That's what I was thinking

Answer 6

then, you are thinking correct :)
any more doubts ?

Answer 7

Well. All you have to do is look at the denominator, right?

Answer 8

and to set it to 0, yes, right.

Answer 9

Unless you can factor the top
EX.
(x^2-9)/x-3
Then it looks like there is a vertical asymptote at x=3, but in reality, it's only a hole at x=3 because you can factor the top and simplify it to cancel the x-3

Answer 10

So what would you do then??

Answer 11

(x^2-9)/x-3
[(x+3)(x-3)]/x-3
(x+3) the x-3 cancels out on the numerator and the denominator, but since the original function had x-3 on the denominator, there is a hole there

Answer 12

So it is discontinuous?

Answer 13

yup!

Answer 14

at that point

Answer 15

Ohh h :o